Question: $f(x) = 7x^{2}-3x+3$ $g(t) = -3t^{3}-7t^{2}-2t-1-2(f(t))$ $h(t) = -4t-4(g(t))$ $ h(g(0)) = {?} $
First, let's solve for the value of the inner function, $g(0)$ . Then we'll know what to plug into the outer function. $g(0) = -3(0^{3})-7(0^{2})+(-2)(0)-1-2(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = 7(0^{2})+(-3)(0)+3$ $f(0) = 3$ That means $g(0) = -3(0^{3})-7(0^{2})+(-2)(0)-1+(-2)(3)$ $g(0) = -7$ Now we know that $g(0) = -7$ . Let's solve for $h(g(0))$ , which is $h(-7)$ $h(-7) = (-4)(-7)-4(g(-7))$ To solve for the value of $h$ , we need to solve for the value of $g(-7)$ $g(-7) = -3(-7)^{3}-7(-7)^{2}+(-2)(-7)-1-2(f(-7))$ To solve for the value of $g$ , we need to solve for the value of $f(-7)$ $f(-7) = 7(-7)^{2}+(-3)(-7)+3$ $f(-7) = 367$ That means $g(-7) = -3(-7)^{3}-7(-7)^{2}+(-2)(-7)-1+(-2)(367)$ $g(-7) = -35$ That means $h(-7) = (-4)(-7)+(-4)(-35)$ $h(-7) = 168$